[英]Regex to replace a word that is not enclosed between tags (non-html)
有限的正則表達式經驗,我正在使用preg_replace在PHP中工作。
我想替換不在[no-glossary] ... [/ no-glossary]標簽之間的指定“單詞”。 如果它們不是“單詞”和標簽之間的空格,或者如果它們是“單詞”之后的空格,那么我的表達有效,但是如果我在單詞之前放置了一個空格(已經過了)它就失敗了!
這些工作:
$html = '<p>Do not replace [no-glossary]this[/no-glossary] replace this.</p>';
$html = '<p>Do not replace [no-glossary]this [/no-glossary] replace this.</p>';
這不是:
$html = '<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>';
使用的模式逐個解釋
/ - find
(?<!\[no-glossary\]) - Not after the [no-glossary] tag
[ ]* - Followed by 0 or more spaces (I think this is the problem)
\b(this)\b - The word "this" between word boundaries
[ ]* - Followed by 0 or more spaces
(?!\[\/no-glossary\]) - Not before the [/no-glossary] tag
/
這是代碼:
$pattern = "/(?<!\[no-glossary\])[ ]*\b(this)\b[ ]*(?!\[\/no-glossary\])/";
$html = '<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>';
$html = preg_replace($pattern, "that", $html);
print $html;
輸出:
<p>Do not change [no-glossary] that [/no-glossary] changethat.</p>
問題:
抓住白色空間:
$subject = <<<LOD
<p>Do not replace [no-glossary]this[/no-glossary] replace this.</p>
<p>Do not replace [no-glossary]this [/no-glossary] replace this.</p>
<p>Do not replace [no-glossary] this[/no-glossary] replace this.</p>
<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>
LOD;
$pattern = '`(?<!\[no-glossary])( *+)\bthis\b( *+)(?!\[/no-glossary])`';
echo $subject.'<br/>';
echo preg_replace($pattern,"$1rabbit$2",$subject); ?>
在使用RegEx模式玩一點之后,我發現Regex PCRE引擎有一些限制,所以我從另一個角度來看問題:
[no-glossary] this [/no-glossary]
和this
。 這可以使用preg_replace_callback()
來完成:
PHP 5.3+要求 :
$pattern = "/\[no-glossary\][ ]*\bthis\b[ ]*\[\/no-glossary\]|this/";
$html = '<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>';
$html = preg_replace_callback($pattern, function($match){
if($match[0] == 'this'){
return('that');
}else{
return($match[0]);
}
}, $html);
print $html;
如果您沒有運行PHP 5.3+:
$pattern = "/\[no-glossary\][ ]*\bthis\b[ ]*\[\/no-glossary\]|this/";
$html = '<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>';
$html = preg_replace_callback($pattern, 'replace_function', $html);
function replace_function($match){
if($match[0] == 'this'){
return('that');
}else{
return($match[0]);
}
}
print $html;
動態:
$tag = 'no-glossary';
$find = 'this';
$replace = 'that';
$pattern = "/\[$tag\][ ]*\b$find\b[ ]*\[\/$tag\]|$find/";
$html = '<p>Do not replace [no-glossary] this [/no-glossary] replace this.</p>';
$html = preg_replace_callback($pattern, function($match) use($find, $replace){
if($match[0] == $find){
return($replace);
}else{
return($match[0]);
}
}, $html);
print $html;
試試這個: ([^\\[\\]])this([^\\[\\]])
當然,你需要在'this'這個詞上應用你真正需要的東西。
試試這個:
\b(this)\b(?!(?:(?!\[no-glossary\]).)*?\[/no-glossary\])
這樣就可以排除更換this
如果它后跟[/no-glossary]
除非遇到[no-glossary]
第一。
試試這個它只取代this
詞
$pattern = '%([^\da-zA-Z]+)this([^\da-zA-Z]+)%si';
$html = '<p>Do not replace [no-glossary]this sdf[/no-glossary] thisreplace<p> replacethis.</p>replace this.</p>';
function Replace1($M){
//print_r($M);
return $M[1]."that".$M[2];
}
$html = preg_replace_callback($pattern,"Replace1",$html);
print $html;
輸出:
<p>Do not replace [no-glossary]that sdf[/no-glossary] thisreplace<p> replacethis.</p>replace that.</p>
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